how to calculate the weight of solute req'd to prepare 1 liter of 1 normal of:

Question

a.) LiOH
b.) phosphoric acid
my work:
formula= gew/L or w/ewxL
i get their mw then divide it to their H/OH so i can get EW then multiply it to 1.
a.) 24 g/mol/1
b.) 98g/mol/3
i don't know what's next to find to be able to follow the formula. how to find the weight!! please help me!

Answer 1

You're there.
Since 24/1 is the EW for LiOH, then 24 g LiOH in 1L solution will be 1 N. (also 1 M)
b. Same for H3PO4. Dissolve 98/3 g H3PO4 in a little water and make to the 1L mark with water to prepare a 1 N solution of H3PO4. Technically you need the reaction being used in H3PO4 since there COULD be 3 different equivalent weights depending upon the number of H ions involved but USUALLY in these questions the maximum is used. So 98/3 is the g required. (By the way that makes a 0.333 M solution)

Answer 2

what am i going to do next? please heelp.